Proof:

Outline

We prove that there is some p such that any generic subset of P including p also includes κ of the pi. Then, since Q⁢[G] satisfies the κ chain condition, two of the corresponding q^i must be compatible. Then, since G is directed, there is some p stronger than any of these which forces this to be true, and therefore makes two elements of S compatible.

Let S=⟨pi,q^i⟩i<κ⊆P*Q.

Claim: There is some p∈P such that p⊩|{i∣pi∈G^}|=κ

(Note: G^={⟨p,p⟩∣p∈P}, hence G^⁢[G]=G)

If no p forces this then every p forces that it is not true, and therefore ⊩P|{i∣pi∈G}|≤κ. Since κ is regular, this means that for any generic G⊆P, {i∣pi∈G} is bounded. For each G, let f⁢(G) be the least α such that β<α implies that there is some γ>β such that pγ∈G. Define B={α∣α=f⁢(G)} for some G.

Claim: |B|<κ

If α∈B then there is some pα∈P such that p⊩f⁢(G^)=α, and if α,β∈B then pα must be incompatible with pβ. Since P satisfies the κ chain condition, it follows that |B|<κ.

Since κ is regular, α=sub⁡(B)<κ. But obviously pα+1⊩pα+1∈G^. This is a contradiction, so we conclude that there must be some p such that p⊩|{i∣pi∈G^}|=κ.

If G⊆P is any generic subset containing p then A={q^i⁢[G]∣pi∈G} must have cardinalityκ. Since Q⁢[G] satisfies the κ chain condition, there exist i,j<κ such that pi,pj∈G and there is some q^⁢[G]∈Q⁢[G] such that q^⁢[G]≤q^i⁢[G],q^j⁢[G]. Then since G is directed, there is some p′∈G such that p′≤pi,pj,p and p′⊩q^⁢[G]≤q^1⁢[G],q^2⁢[G]. So ⟨p′,q^⟩≤⟨pi,q^i⟩,⟨pj,q^j⟩.